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p 1 b 0 g 0 p 0 p 2 g 1 b 1 p 3 Suppose that the two boys b 0 and b 1 and the two girls g 0 and g 1 are not involved in any other triples. (The four pets p 0 , . . . , p 3 will of course belong to other triples as well; for otherwise the instance would trivially have no solution.) Then any matching must contain either the two triples (b 0 , g 1 , p 0 ), (b 1 , g 0 , p 2 ) or the two triples (b 0 , g 0 , p 1 ), (b 1 , g 1 , p 3 ), because these are the only ways in which these two boys and girls can find any match. Therefore, this “gadget” has two possible states: it behaves like a Boolean variable! To then transform an instance of 3SAT to one of 3D MATCHING, we start by creating a copy of the preceding gadget for each variable x. Call the resulting nodes p x1 , b x0 , g x1 , and so on. The intended interpretation is that boy b x0 is matched with girl g x1 if x = true, and with girl g x0 if x = false. Next we must create triples that somehow mimic clauses. For each clause, say c = (x∨y∨z), introduce a new boy b c and a new girl g c . They will be involved in three triples, one for each literal in the clause. And the pets in these triples must reflect the three ways whereby the clause can be satisfied: (1) x = true, (2) y = false, (3) z = true. For (1), we have the triple (b c , g c , p x1 ), where p x1 is the pet p 1 in the gadget for x. Here is why we chose p 1 : if x = true, then b x0 is matched with g x1 and b x1 with g x0 , and so pets p x0 and p x2 are taken. In which case b c and g c can be matched with p x1 . But if x = false, then p x1 and p x3 are taken, and so g c and b c cannot be accommodated this way. We do the same thing for the other two literals of the clause, which yield triples involving b c and g c with either p y0 or p y2 (for the negated variable y) and with either p z1 or p z3 (for variable z). We have to make sure that for every occurrence of a literal in a clause c there is a different pet to match with b c and g c . But this is easy: by an earlier reduction we can assume that no literal appears more than twice, and so each variable gadget has enough pets, two for negated occurrences and two for unnegated. The reduction now seems complete: from any matching we can recover a satisfying truth assignment by simply looking at each variable gadget and seeing with which girl b x0 was matched. And from any satisfying truth assignment we can match the gadget corresponding to each variable x so that triples (b x0 , g x1 , p x0 ) and (b x1 , g x0 , p x2 ) are chosen if x = true and triples (b x0 , g x0 , p x1 ) and (b x1 , g x1 , p x3 ) are chosen if x = false; and for each clause c match b c and g c with the pet that corresponds to one of its satisfying literals. But one last problem remains: in the matching defined at the end of the last paragraph, some pets may be left unmatched. In fact, if there are n variables and m clauses, then exactly 252
2n − m pets will be left unmatched (you can check that this number is sure to be positive, because we have at most three occurrences of every variable, and at least two literals in every clause). But this is easy to fix: Add 2n − m new boy-girl couples that are “generic animallovers,” and match them by triples with all the pets! 3D MATCHING−→ZOE Recall that in ZOE we are given an m × n matrix A with 0 − 1 entries, and we must find a 0 − 1 vector x = (x 1 , . . . , x n ) such that the m equations Ax = 1 are satisfied, where by 1 we denote the column vector of all 1’s. How can we express the 3D MATCHING problem in this framework? ZOE and ILP are very useful problems precisely because they provide a format in which many combinatorial problems can be expressed. In such a formulation we think of the 0 − 1 variables as describing a solution, and we write equations expressing the constraints of the problem. For example, here is how we express an instance of 3D MATCHING (m boys, m girls, m pets, and n boy-girl-pet triples) in the language of ZOE. We have 0 − 1 variables x 1 , . . . , x n , one per triple, where x i = 1 means that the ith triple is chosen for the matching, and x i = 0 means that it is not chosen. Now all we have to do is write equations stating that the solution described by the x i ’s is a legitimate matching. For each boy (or girl, or pet), suppose that the triples containing him (or her, or it) are those numbered j 1 , j 2 , . . . , j k ; the appropriate equation is then x j1 + x j2 + · · · + x jk = 1, which states that exactly one of these triples must be included in the matching. For example, here is the A matrix for an instance of 3D MATCHING we saw earlier. Al Bob Chet Alice Beatrice Carol ⎛ ⎞ 1 0 0 0 0 0 0 0 1 1 0 1 1 0 0 1 0 0 0 1 A = 0 1 0 0 0 0 0 1 1 0 ⎜1 0 0 0 1 ⎟ ⎝0 0 1 1 0⎠ 0 1 0 0 0 Armadillo Bobcat Canary The five columns of A correspond to the five triples, while the nine rows are for Al, Bob, Chet, Alice, Beatrice, Carol, Armadillo, Bobcat, and Canary, respectively. It is straightforward to argue that solutions to the two instances translate back and forth. 253
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Algorithms Copyright c○2006 S. Da
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4 Paths in graphs 109 4.1 Distances
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List of boxes Bases and logs . . .
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Preface This book evolved over the
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Chapter 0 Prologue Look around you.
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The first question is moot here, as
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for addition, which works on one di
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4. Likewise, any polynomial dominat
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and in general ( Fn ) = F n+1 ( ) n
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Bases and logs Naturally, there is
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Figure 1.1 Multiplication à la Fra
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Figure 1.3 Addition modulo 8. 6 0 0
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Figure 1.4 Modular exponentiation.
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Figure 1.6 A simple extension of Eu
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We say x is the multiplicative inve
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Figure 1.7 An algorithm for testing
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Figure 1.8 An algorithm for testing
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Randomized algorithms: a virtual ch
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Alice’s encryption function is th
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Figure 1.9 RSA. Bob chooses his pub
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There is nothing inherently wrong w
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Exercises 1.1. Show that in any bas
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the depth of the circuit or the lon
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(c) Give an example of positive int
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x = x L x R = 2 n/2 x L + x R y = y
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Figure 2.2 Divide-and-conquer integ
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Binary search The ultimate divide-a
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An n log n lower bound for sorting
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The three sublists S L , S v , and
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to be the best running time possibl
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qp t AB CD EF GH IJ KL MN OP QR Why
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The equivalence of the two polynomi
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Figure 2.6 The complex roots of uni
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A matrix reformulation To get a cle
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The orthogonality property can be s
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FFT n (input: a 0 , . . . , a n−1
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The slow spread of a fast algorithm
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(e) T (n) = 8T (n/2) + n 3 (f) T (n
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2.21. Mean and median. One of the m
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(c) In fact, squaring matrices is n
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Chapter 3 Decompositions of graphs
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Therefore, each edge appears in exa
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Figure 3.3. 1 The previsit and post
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Figure 3.6 (a) A 12-node graph. (b)
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Tree edges are actually part of the
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the same thing. Since a dag is line
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Figure 3.10 The reverse of the grap
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Exercises 3.1. Perform a depth-firs
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(a) Model this as a graph problem:
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For instance, in the graph below (w
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3.30. On page 97, we defined the bi
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Chapter 4 Paths in graphs 4.1 Dista
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Figure 4.4 The result of breadth-fi
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Figure 4.6 Breaking edges into unit
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into account. This viewpoint gives
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§¦ ¥¤ £¢ ©¨ #"
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Which heap is best? The running tim
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Figure 4.11 (a) A binary heap with
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Figure 4.13 The Bellman-Ford algori
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Figure 4.15 A single-source shortes
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Input: Undirected graph G = (V, E)
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4.16. Section 4.5.2 describes a way
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Let r ∗ be the maximum ratio achi
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Chapter 5 Greedy algorithms A game
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Trees A tree is an undirected graph
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Figure 5.3 The cut property at work
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eturn x As can be expected, makeset
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Figure 5.7 The effect of path compr
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A randomized algorithm for minimum
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Figure 5.9 Top: Prim’s minimum sp
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Figure 5.10 A prefix-free encoding.
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149
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5.3 Horn formulas In order to displ
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Figure 5.11 (a) Eleven towns. (b) T
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Exercises 5.1. Consider the followi
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(a) Code: {0, 10, 11} (b) Code: {0,
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Input: Undirected graph G = (V, E);
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Chapter 6 Dynamic programming In th
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Figure 6.2 The dag of increasing su
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Programming? The origin of the term
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Figure 6.4 (a) The table of subprob
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Common subproblems Finding the righ
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6.4 Knapsack During a robbery, a bu
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Memoization In dynamic programming,
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Figure 6.7 (a) ((A × B) × C) × D
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ecause it leads right back to the O
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d ij . We must pick the best such i
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Figure 6.11 I(u) is the size of the
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6.8. Given two strings x = x 1 x 2
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Figure 6.12 Two binary search trees
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6.26. Sequence alignment. When a ne
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Chapter 7 Linear programming and re
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It is a general rule of linear prog
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the feasible region. The point of f
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Figure 7.3 A communications network
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Reductions Sometimes a computationa
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Matrix-vector notation A linear fun
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Page 205 and 206: Figure 7.6 Continued Current Flow R
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Page 209 and 210: Figure 7.10 A generic primal LP in
Page 211 and 212: Now suppose the two of them play th
Page 213 and 214: In LP form, this is min w −3y 1 +
Page 215 and 216: 7.6.2 The algorithm On each iterati
Page 217 and 218: Figure 7.13 Simplex in action. Init
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Page 221 and 222: Gaussian elimination Under our alge
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Page 227 and 228: 7.12. For the linear program max x
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Page 231 and 232: 7.28. A linear program for shortest
Page 233 and 234: Chapter 8 NP-complete problems 8.1
Page 235 and 236: Satisfiability SATISFIABILITY, or S
Page 237 and 238: cost as the budget. Fine, but how d
Page 239 and 240: Figure 8.3 What is the smallest cut
Page 241 and 242: Figure 8.4 A more elaborate matchma
Page 243 and 244: 8.2 NP-complete problems Hard probl
Page 245 and 246: I. These two translation procedures
Page 247 and 248: Figure 8.7 Reductions between searc
Page 249 and 250: Figure 8.8 The graph corresponding
Page 251: Figure 8.9 S is a vertex cover if a
Page 255 and 256: Figure 8.10 Rudrata cycle with pair
Page 257 and 258: est of the graph as shown in Figure
Page 259 and 260: RUDRATA CYCLE−→TSP Given a grap
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Page 263 and 264: Exercises 8.1. Optimization versus
Page 265 and 266: Input: An undirected graph G = (V,
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Page 269 and 270: Chapter 9 Coping with NP-completene
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Page 273 and 274: follow the same pattern. As before,
Page 275 and 276: 9.2 Approximation algorithms In an
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Page 279 and 280: 9.2.3 TSP The triangle inequality p
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Page 285 and 286: Figure 9.8 Local search. Figure 9.7
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Page 289 and 290: Figure 9.10 The search space for a
Page 291 and 292: Exercises 9.1. In the backtracking
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Page 295 and 296: Chapter 10 Quantum algorithms This
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Page 299 and 300: Figure 10.3 A quantum algorithm tak
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The Fourier transform of a periodic
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Yet another basic gate, the control
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10.6 Factoring as periodicity We ha
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Figure 10.6 Quantum factoring. 0 QF
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Exercises 10.1. ∣ ψ 〉 = 1 √2
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Historical notes and further readin
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Index O(·), 13 Ω(·), 14 Θ(·),
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interior-point method, 217 interpol
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